Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
422 views
in Technique[技术] by (71.8m points)

ios - NSPredicate filtered by year moth day

I use Core Data for storing my data model objects. Each object has NSDate property.

NSDate property has format like below:

2013-03-18 12:50:31 +0000

I need to create predicate that will fetch my objects just by this value 2013-03-18 without time.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

If your dates are stored as actual dates then you should use that to your advantage and not fiddle with formats. You can simply create a predicate that checks that if the dates is between two dates (with times). The first date is your date with the time 00:00:00 and the second date is one day after that.

// Create your date (without the time)
NSDateComponents *yourDate = [NSDateComponents new];
yourDate.calendar = [NSCalendar currentCalendar];
yourDate.year  = 2013;
yourDate.month = 3;
yourDate.day   = 18;
NSDate *startDate = [yourDate date];

// Add one day to the previous date. Note that  1 day != 24 h
NSDateComponents *oneDay = [NSDateComponents new];
oneDay.day = 1;
// one day after begin date
NSDate *endDate = [[NSCalendar currentCalendar] dateByAddingComponents:oneDay 
                                                                toDate:startDate
                                                               options:0];

// Predicate for all dates between startDate and endDate
NSPredicate *dateThatAreOnThatDay = 
    [NSPredicate predicateWithFormat:@"(date >= %@) AND (date < %@)", 
                                     startDate, 
                                     endDate]];

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...